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Tresset [83]
3 years ago
11

Please help I think I know the answers but I’m not positive if it’s correct

Mathematics
1 answer:
kvasek [131]3 years ago
3 0
CDEF is the right answer
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PLEASE HELP ME PLEASE
Sergio [31]

Answer:

A

Step-by-step explanation:

Given

\frac{3}{4} (- 12x + 4) + \frac{1}{2} (6x + 8) ← distribute parenthesis

= - 9x + 3 + 3x + 4 ← collect like terms

= - 6x + 7 → A

4 0
3 years ago
Mrs. Gordon bought 10 kg of apples for $16.50. What is the price per pound of apples?
natulia [17]

Answer:

1.65 per kg

Step-by-step explanation:

4 0
3 years ago
Graph the hyperbola using the transverse axis, vertices, and co-vertices:
Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola

Read more about hyperbola at:

brainly.com/question/26250569

#SPJ1

4 0
1 year ago
Hey can you please help me posted picture of question
AnnZ [28]
I think choice a would be the correct answer. 90.
8 0
3 years ago
Pleaase help me because i really need it
MrMuchimi

Answer:

B and C are both correct

Step-by-step explanation:

18/6 = 3

h¹⁰/h¹² = h⁽¹⁰⁻¹²⁾ = h⁻² = 1/h²

the a⁻⁴ term remains unchanged and = 1/a⁴

3a⁻⁴h⁻² = \frac{3}{a^4h^2}

5 0
2 years ago
Read 2 more answers
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